a) phương trình \(x^3-3x^2+1\) có 3 nghiệm thực phân biệt là a,b,c(đề bài). Áp dụng Định lí Vi-ét cho đa thức bậc 3 ta có:\(\left\{{}\begin{matrix}a+b+c=3\\ab+bc+ac=0\\a.b.c=-1\end{matrix}\right.\)

ta có

      a+b+c=3

<=>\(\left(a+b+c\right)^2=9\)

<=>\(a^2+b^2+c^2+2ab+2bc+2ac=9\)

<=>\(a^2+b^2+c^2=9\)

<=>\(\left(a^2+b^2+c^2\right)^2=81\)

<=>\(a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=81\)(1)

ta có ab+bc+ac=0

   <=>\(\left(ab+bc+ac\right)^2=0\)

   <=>\(a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=0\)

   <=>\(a^2b^2+b^2c^2+a^2c^2-2.1.3=0\)

   <=>\(a^2b^2+b^2c^2+a^2c^2=6\)(2)

Thay (2) vào (1) ta có \(a^4+b^4+c^4+2.6=81\)

                                <=>\(a^4+b^4+c^4=69\)

b) \(\dfrac{a+1}{\left(b+c\right)\left(1-a\right)+1}=\dfrac{a+1}{\left(3-a\right)\left(1-a\right)+1}=\dfrac{a+1}{3+a^2-4a+1}=\dfrac{a+1}{a^2-4a+4}=\dfrac{a+1}{\left(a-2\right)^2}\)

cmtt =>\(B=\dfrac{a+1}{\left(a-2\right)^2}+\dfrac{b+1}{\left(b-2\right)^2}+\dfrac{c+1}{\left(c-2\right)^2}\)=\(\dfrac{1}{a-2}+\dfrac{1}{b-2}+\dfrac{1}{c-2}+3\left[\dfrac{1}{\left(a-2\right)^2}+\dfrac{1}{\left(b-2\right)^2}+\dfrac{1}{\left(c-2\right)^2}\right]\)=\(\dfrac{3\left[\left(a-2\right)\left(b-2\right)\right]^2+3\left[\left(b-2\right)\left(c-a\right)\right]^2+3\left[\left(c-2\right)\left(a-2\right)\right]^2}{\left[\left(a-2\right)\left(b-2\right)\left(c-2\right)\right]^2}\)

đặt t=(a-2)(b-2);u=(b-2)(c-2);v=(c-2)(a-2)     =>t+u+v=0

B thành \(\dfrac{3\left(t^2+u^2+v^2\right)}{t.u.v}\) bạn biến đổi để xuất hiện t+u+v

=>B=\(\dfrac{3\left(t+u+v\right)^2-6\left(t.u+u.v+t.v\right)}{t.u.v}=\dfrac{-6.\left(a-2\right)\left(b-2\right)\left(c-2\right)\left(a-2+b-2+c-2\right)}{t.u.v}=\dfrac{18}{\left(a-2\right)\left(b-2\right)\left(c-2\right)}\)

(a-2)(b-2)(c-2)= abc-2(ab+bc+ac)+4(a+b+c)-8=12-9=3

Vậy B=3

2

a

\(x+y+z=0\)

\(\Rightarrow x+y=-z\)

\(\Rightarrow\left(x+y\right)^3=\left(-z\right)^3\)

\(\Rightarrow x^3+y^3+3x^2y+3xy^2=-z^3\)

\(\Rightarrow x^3+y^3+z^3=3xy\left(x+y\right)=3xyz\)

b

Đặt \(a-b=x;b-c=y;c-a=z\Rightarrow x+y+z=0\)

Ta có bài toán mới:Cho \(x+y+z=0\).Phân tích đa thức thành nhân tử:\(x^3+y^3+z^3\)

Áp dụng kết quả câu a ta được:

\(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

bạn viết có thánh đọc ra á :v

Bạn viết như vậy vẫn nhìn đc nhưng nhìn hơi khó

a: \(\left(a-2b\right)^2-4b^2\)

\(=\left(a-2b\right)^2-\left(2b\right)^2\)

=(a-2b-2b)(a-2b+2b)

=a(a-4b)

b: \(\left(a-b\right)^2-c^2=\left(a-b-c\right)\left(a-b+c\right)\)

c: \(\left(a+b\right)^2-4=\left(a+b\right)^2-2^2=\left(a+b+2\right)\left(a+b-2\right)\)

d: \(\left(a+3b\right)^2-9b^2\)

\(=\left(a+3b\right)^2-\left(3b\right)^2\)

=(a+3b-3b)(a+3b+3b)=a(a+6b)

e: \(\left(x-3\right)^3-27\)

\(=\left(x-3-3\right)\left\lbrack\left(x-3\right)^2+3\left(x-3\right)+9\right\rbrack\)

\(=\left(x-6\right)\left(x^2-6x+9+3x-9+9\right)=\left(x-6\right)\left(x^2-3x+9\right)\)

f: \(\left(x+1\right)^3-125\)

\(=\left(x+1-5\right)\left\lbrack\left(x+1\right)^2+5\left(x+1\right)+25\right\rbrack\)

\(=\left(x-4\right)\left(x^2+2x+1+5x+5+25\right)=\left(x-4\right)\left(x^2+7x+31\right)\)

dấu ^ là mũ nha mn

vại

fdvfdverberrgtrgrgg

1) \(2xy^3-6x^2+10xy\)

\(=2x.y^3-2x.3x+2x.5y\)

\(=2x\left(y^3-3x+5y\right)\)

\(=2x[y\left(y^2-5\right)-3x]\)

2) \(a^6-a^5-2a^3+2a^2\)

\(=\left(a^6-a^5\right)-\left(2a^3-2a^2\right)\)

\(=\left(a^5.a-a^5.1\right)-\left(2a^2.a-2a^2.1\right)\)

\(=a^5\left(a-1\right)-2a^2\left(a-1\right)\)

\(=\left(a^5-2a^2\right)\left(a-1\right)\)

\(=a^2\left(a^3-2\right)\left(a-1\right)\)